LXXXIV.1 (1998)
On the average number of
unitary factors of finite abelian groups
by
J. Wu (Nancy)
1. Introduction. Let X be the semigroup of all finite abelian groups with respect to the direct product ⊗ and let E
0be the identity of X. For G ∈ X and H ∈ X, we use (G, H) to denote the group of maximal order in X which is simultaneously a direct factor of G and H. We say that G and H are relatively prime if (G, H) = E
0. A direct factor D of G is called unitary if D ⊗ E = G and (D, E) = E
0. The number of unitary factors of G is denoted by t(G). In 1960, Cohen [2] proved
(1.1) X
|G|≤x
t(G) = A
1x log x + A
2x + O( √
x log x),
where the summation is over all G in X of order |G| ≤ x and the A
jare some effective constants. After a study on Dirichlet’s series associated with t(G), Kr¨atzel [7] found a connection between (1.1) and the following three- dimensional divisor problem:
(1.2) X
n1n2n23≤x
1 = B
1x log x + B
2x + B
3√
x + ∆(1, 1, 2; x),
where the B
jare some effective constants and ∆(1, 1, 2; x) is an error term. Using exponential sum techniques, he showed that ∆(1, 1, 2; x) x
11/29(log x)
2, which implies
(1.3) X
|G|≤x
t(G) = A
1x log x + A
2x + A
3√ x + ∆(x)
with ∆(x) x
11/29(log x)
2. This estimate was improved to ∆(x) x
3/8(log x)
4by Schmidt [11], then to ∆(x)
εx
77/208+εby Liu [9] and to
∆(x)
εx
29/80+εby Liu [10], where ε denotes an arbitrarily small positive number.
1991 Mathematics Subject Classification: 11L07, 11N45.
[17]
In this paper, we give a better bound.
Theorem 1. For any ε > 0, we have
∆(1, 1, 2; x)
εx
47/131+εand ∆(x)
εx
47/131+ε.
For comparison, we have 29/80 = 0.3625 and 47/131 ≈ 0.3587. From [10], we know that in the proof of Theorem 1 the most difficult part is to estimate the exponential sums of type
(1.4) X
h∼H
a
hX
n1∼N1
X
n2∼N2
e(xh/(n
21n
2)),
where |a
h| ≤ 1, e(t) := e
2πitand the notation h ∼ H means cH < h ≤ c
0H with some positive unspecified constants c, c
0. Liu [10] has treated (1.4) by combining Fouvry–Iwaniec’s method [3] and Kolesnik’s method [6].
We notice that via van der Corput’s B-process the sum (1.4) can be transformed into bilinear exponential sums of type I,
T (M, N ) := X
m∼M
X
n∈I(m)
ϕ
me
X m
αn
βM
αN
β,
where I(m) is a subinterval of [N, 2N ]. Using the classical A, B process and the well known AB theorem of Kolesnik (see Theorem 1 of [6] and Lemma 1.5 of [8]) we shall prove an estimate for T (M, N ) (see Theorem 3 below). In addition we also use an idea of Jia ([5], Lemma 13) and Liu ([8], Lemma 2.4) to investigate bilinear exponential sums of type II,
S(M, N ) := X
m∼M
X
n∼N
ϕ
mψ
ne
X m
αn
βM
αN
β.
Baker and Harman have simplified Jia–Liu’s argument to obtain a slightly more general estimate for S(M, N ) (see Theorem 2 of [1]) than those of Jia and Liu. But all such results contain some restrictions on (X, M, N ) and the number of terms is relatively large; this is not convenient in applications.
Our result (see Theorem 2 below) essentially has the same power as their estimates, but it is without restriction, more general and simpler in form.
Finally, it is worth indicating that we also need Theorem 7 of [12] and Lemma 2.3 of [13] for the proof of Theorem 1.
2. Estimates for exponential sums. We first prove two estimates for S = S(M, N ), defined as in Section 1. In the sequel, the letter ε
0denotes a suitably small positive number (depending on α, β and α
jat most).
Theorem 2. Let α, β ∈ R with αβ(α − 1)(β − 1) 6= 0, X > 0, M ≥ 1,
N ≥ 1, |ϕ
m| ≤ 1, |ψ
n| ≤ 1 and L := log(2 + XM N ). If (κ, λ) is an exponent
pair , then
S(M, N ) {(X
2+4κM
8+10κN
9+11κ+λ)
1/(12+16κ)(2.1)
+ X
1/6M
2/3N
3/4+λ/(12+12κ)+ (XM
3N
4)
1/5+ (XM
7N
10)
1/11+ M
2/3N
11/12+λ/(12+12κ)+ M N
1/2+ (X
−1M
14N
23)
1/22+ X
−1/2M N }L
2, S(M, N ) {(XM
3N
4)
1/5+ (X
4M
10N
11)
1/16+ (XM
7N
10)
1/11(2.2)
+ M N
1/2+ (X
−1M
14N
23)
1/22+ X
−1/2M N }L
2. P r o o f. We begin in the same way as Jia [5], Liu [8], Baker and Harman [1]. Without loss of generality, we suppose that β > 0 and L is sufficiently large. Let Q ∈ [L, N/L] be a parameter to be chosen later. By Cauchy’s inequality and a “Weyl shift” ([4], Lemma 2.5), we have
|S|
2(M N )
2Q + M
3/2N
Q
X
1≤|q1|<Q
1 − |q
1| Q
X
n+q1,n∼N
ψ
n+q1ψ
nX
m∼M
m
−1/2e(Am
αt),
where t = t(n, q
1) := (n+q
1)
β−n
βand A := X/(M
αN
β). Splitting the range of q
1into dyadic intervals and removing 1 − q
1/Q by partial summation, we get
(2.3) |S|
2(M N )
2Q
−1+ LM
3/2N Q
−1max
1≤Q1≤Q
|S(Q
1)|, where
S(Q
1) := X
q1∼Q1
X
n+q1,n∼N
ψ
n+q1ψ
nX
m∼M
m
−1/2e(Am
αt).
If X(M N )
−1Q
1≥ ε
0, by Lemma 2.2 of [12] we can transform the inner- most sum to a sum over l and then using Lemma 2.3 of [12] with n = m we can estimate the corresponding error term. As a result, we obtain
S(Q
1) X
q1∼Q1
X
n+q1,n∼N
ψ
n+q1ψ
nX
l∈I
l
−1/2e(e α(At)
γl
1−γ) + {(XM
−1N
−1Q
31)
1/2+ M
−1/2N Q
1+ (X
−1M N Q
1)
1/2+ (X
−2M N
4)
1/2}L,
where γ := 1/(1−α), e α = |1−α|·|α|
α/(1−α), I := [c
1AM
α−1|t|, c
2AM
α−1|t|]
and c
j= c
j(α) are some constants. Exchanging the order of summation and
estimating the sum over l trivially, we find, for some l X(M N )
−1Q
1, the
inequality
S(Q
1) (XM
−1N
−1Q
1)
1/2X X
(n,q1)∈D1(l)
ψ
n+q1ψ
ne(e α(At)
γl
1−γ) (2.4)
+ {(XM
−1N
−1Q
31)
1/2+ M
−1/2N Q
1+ (X
−1M N Q
1)
1/2+ (X
−2M N
4)
1/2}L,
where D
1(l) is a suitable subregion of {(n, q
1) : n ∼ N, q
1∼ Q
1}. Let S
1(Q
1) be the double sums on the right-hand side of (2.4). Using Lem- ma 2.6 of [12] to relax the range of q
1, we see that there exists a real number θ independent of (n, q
1) such that
S
1(Q
1) L X
n∼N
X
q1∼Q1
ψ
n+q1e(θq
1)e(e α(At)
γl
1−γ) .
If L ≤ Q
1≤ Q, using again Cauchy’s inequality and a “Weyl shift” with Q
2≤ ε
0√
Q
1yields
|S
1(Q
1)/L|
2(N Q
1)
2Q
−12+ N Q
1Q
−12X
1≤q2≤Q2
|S
2(q
1, q
2)|, where
S
2(q
1, q
2) := X
n∼N
X
q1+q2,q1∼Q1
ψ
n+q1+q2ψ
n+q1e(t
1(n, q
1, q
2))
and t
1(n, q
1, q
2) := e αA
γl
1−γ{t(n, q
1+ q
2)
γ− t(n, q
1)
γ}. Writing n
0:= n + q
1, exchanging the order of summation and using Lemma 2.6 of [12], we can deduce
S
2(q
1, q
2) = X X
(n0,q1)∈D2
ψ
n0+q2ψ
n0e(t
1(n
0− q
1, q
1, q
2))
L X
n0∼N
X
q1∼Q1
e(θ
0q
1)e(T (n
0, q
1, q
2)) ,
where T (n
0, q
1, q
2) := t
1(n
0−q
1, q
1, q
2), D
2is a suitable subregion of {(n
0, q
1) : n
0∼ N, q
1∼ Q
1} and θ
0is a real number independent of (n
0, q
1). A final application of Cauchy’s inequality and a “Weyl shift” with Q
3= Q
22gives
|S
2(q
1, q
2)/L|
2(N Q
1)
2Q
−13+ N Q
1Q
−13X
1≤q3≤Q3
X
q1∼Q1
|S
3(q
1, q
2, q
3)|, where S
3(q
1, q
2, q
3) := P
n0∼N
e(f (n
0)) and f (n
0) := T (n
0, q
1, q
2) − T (n
0, q
1+ q
3, q
2). It is easy to show that f (n
0) satisfies the conditions of exponent pair and f
0(n
0) XN
−2Q
−11q
2q
3(n
0∼ N ). Hence we have
S
3(q
1, q
2, q
3) (XN
−2Q
−11q
2q
3)
κN
λ+ (XN
−2Q
−11q
2q
3)
−1,
which implies
S
1(Q
1) {(X
κN
3−2κ+λQ
4−κ1Q
3κ2)
1/4+ N Q
1Q
−1/22+ (X
−1N
5Q
51Q
−32)
1/4}L
7/4provided Q
1≥ L, Q
2≤ ε
0√
Q
1. By Lemma 2.4(ii) of [12] optimizing Q
2over (0, ε
0√ Q
1] yields
S
1(Q
1) {(X
κN
3+4κ+λQ
4+5κ1)
1/(4+6κ)+ N
3/4+λ/(4+4κ)Q
1+ N Q
3/41+ (X
−2N
10Q
71)
1/8}L
7/4provided Q
1≥ L. In view of the term N Q
3/41L
7/4, this inequality holds trivially when Q
1≤ L. Inserting the preceding estimate in (2.4) yields, for any Q
1∈ [1, Q],
S(Q
1) {(X
2+4κM
−2−3κN
1+κ+λQ
6+8κ1)
1/(4+6κ)+ X
1/2M
−1/2N
1/4+λ/(4+4κ)Q
3/21+ (X
2M
−2N
2Q
51)
1/4+ (X
2M
−4N
6Q
111)
1/8+ (XM
−1N
−1Q
31)
1/2+ M
−1/2N Q
1+ (X
−1M N Q
1)
1/2+ (X
−2M N
4)
1/2}L
7/4=: (E
1+ E
2+ . . . + E
8)L
7/4.
Since E
5≤ E
3and E
6= (E
44E
8)
1/5(M
2Q
1)
−1/10, both E
5and E
6are superfluous. Replacing Q
1by Q and inserting the bound obtained in (2.3), we find, for any Q ∈ [L, N/L],
S {(X
2+4κM
4+6κN
5+7κ+λQ
2+2κ)
1/(8+12κ)(2.5)
+ X
1/4M
1/2N
5/8+λ/(8+8κ)Q
1/4+ (X
2M
4N
6Q)
1/8+ (X
2M
8N
14Q
3)
1/16+ M N Q
−1/2+ (X
−1M
2N
3Q
−1)
1/2}L
11/8,
where we have used the fact that (X
−1M
4N
3Q
−1)
1/4can be absorbed by M N Q
−1/2. In view of M N Q
−1/2, the preceding estimate holds trivially when Q ∈ (0, L].
If X(M N )
−1Q
1≤ ε
0, we can remove m
−1/2by partial summation and then estimate the sum over m by Kuz’min–Landau’s inequality ([4], The- orem 2.1). Hence we see that (2.5) always holds for 0 < Q ≤ N/L. Using Lemma 2.4(ii) of [12] to optimize Q over (0, N/L] yields
S {(X
2+4κM
8+10κN
9+11κ+λ)
1/(12+16κ)+ (X
2κM
8+10κN
11+13κ+λ)
1/(12+16κ)+ X
1/6M
2/3N
3/4+λ/(12+12κ)+ M
2/3N
11/12+λ/(12+12κ)+ (XM
3N
4)
1/5+ (XM
6N
9)
1/10+ (XM
7N
10)
1/11+ (X
−1M
14N
23)
1/22+ M N
1/2+ X
−1/2M N }L
2=: (F
1+ F
2+ . . . + F
10)L
2. Since
F
2= (F
46+3κF
55κ)
1/(6+8κ)N
−κ(1+κ−λ)/((4+4κ)(6+8κ))and F
6= (F
516F
811)
1/27M
−2/135, they are both superfluous. This proves (2.1).
To prove (2.2), we take Q
2= ε
0min{ √
Q
1, (X
−1N
2Q
1)
1/3} such that
|f
0(n
0)| ≤ 1/2 for n
0∼ N . Thus Kuz’min–Landau’s inequality gives S
3(q
1, q
2, q
3) (XN
−2Q
−11q
2q
3)
−1, from which we can deduce, as before, the following inequality:
S {(XM
3N
4)
1/5+ (XM
6N
9)
1/10+ (X
4M
10N
11)
1/16+ (X
2M
10N
13)
1/16+ (XM
7N
10)
1/11+ (X
−1M
14N
23)
1/22+ M N
1/2+ X
−1/2M N + M
1/2N }L
2=: (G
1+ G
2+ . . . + G
9)L
2.
It is not difficult to verify that G
2= (G
161G
116)
1/27M
−2/135, G
4= (G
153G
116)
1/26(M
2N
11)
−1/416, G
9= (G
5G
26)
1/3M
−3/22. Thus G
2, G
4, G
9are superfluous. This completes the proof.
For T = T (M, N ) defined as in Section 1, we have the following result.
Theorem 3. Let α, β ∈ R with αβ(α−1)(β−1)(α+β−1)(2α+β−2) 6= 0, X > 0, M ≥ 1, N ≥ 1, L := log(2 + XM N ), |ϕ
m| ≤ 1 and I(m) be a subinterval of [N, 2N ]. Then
T (M, N ) {(X
5M
10N
8)
1/16+ (X
3M
10N
12)
1/16+ (XM
2N
3)
1/4+ (X
3M
14N
18)
1/22+ (XM
6N
9)
1/10+ (X
7M
30N
24)
1/40+ (XM
5N
5)
1/7+ M N
1/2+ X
−1M N }L
3.
P r o o f. If X ≤ ε
0N , then T X
−1M N by Kuz’min–Landau’s inequal- ity. When X ≥ ε
0N , using (2.3) with ψ
n= 1, we have, for any 1 ≤ Q ≤ ε
0N , (2.6) |T |
2(M N )
2Q
−1+ LM
3/2N Q
−1max
1≤Q1≤Q
|T (Q
1)|, where
T (Q
1) := X
q∼Q1
X
n∈I1(q)
X
m∈J(n,q)
m
−1/2e
X m
αt(n, q) M
αN
β,
t(n, q) := (n + q)
β− n
β, I
1(q) is a subinterval of [N, 2N ] and J(n, q) a
subinterval of [M, 2M ].
If L := X(M N )
−1Q
1≥ ε
0, similarly to (2.4), we can prove, for some l L,
T (Q
1) (XM
−1N
−1Q
1)
1/2X X
(n,q)∈D(l)
e(f (n, q)) + {M
−1/2N Q
1+ (XM
−1N
−1Q
31)
1/2+ (X
−2M N
4)
1/2+ (X
−1M N Q
1)
1/2}L,
where f (n, q) := e α(XQ
1/N )(l/L)
α/(α−1){t(n, q)/(N
β−1Q
1)}
1/(1−α)and D(l) is a suitable subregion of {(n, q) : n ∼ N, q ∼ Q
1}. It is easy to show that f (n, q) satisfies the condition of Lemma 1.5 of [8] (which is a revised form of Theorem 1 of Kolesnik [6]) with A = XN
−1Q
1/(N
β−1Q
1)
1/(1−α),
∆ = Q
1/N . By this lemma with (F, X, Y ) = (XN
−1Q
1, N, Q
1), we obtain the estimate
T (Q
1) {(X
5M
−3N
−2Q
81)
1/6+ (X
3M
−3N
2Q
81)
1/6(2.7)
+ (XM
−1N Q
21)
1/2+ (X
3M
−4N
4Q
111)
1/8+ (XM
−2N
3Q
51)
1/4+ (X
7M
−5N
−6Q
201)
1/10+ (X
2M
−2Q
71)
1/4+ (X
−2M N
4)
1/2+ (X
−1M N Q
1)
1/2}L
4, where we have used the fact that M
−1/2N Q
1+ (XM
−1N
−1Q
31)
1/2can be absorbed by (XM
−2N
3Q
51)
1/4+ (X
2M
−2Q
71)
1/4(in view of the hypothesis X ≥ ε
0N ).
If L ≤ ε
0, the Kuz’min–Landau inequality implies that (2.7) also holds.
Replacing Q
1by Q and inserting into (2.6) yield
|T |
2{(X
5M
6N
4Q
2)
1/6+ (X
3M
6N
8Q
2)
1/6+ (XM
2N
3)
1/2+ (X
3M
8N
12Q
3)
1/8+ (XM
4N
7Q)
1/4+ (X
7M
10N
4Q
10)
1/10+ (X
2M
4N
4Q
3)
1/4+ (M N )
2Q
−1}L
5,
where we have eliminated two superfluous terms X
−1M
2N
3Q
−1and (X
−1M
4N
3Q
−1)
1/2(which can be absorbed by (M N )
2Q
−1). Using Lem- ma 2.4(ii) of [12] to optimize Q over (0, ε
0N ] gives the required result. This concludes the proof.
Next we shall apply Theorems 2 and 3 to treat S
I:= X
m1∼M1
X
m2∼M2
X
m3∼M3
ψ
m2e
X m
α11m
α22m
−α3 2M
1α1M
2α2M
3−α2,
S
II:= X
m1∼M1
X
m2∼M2
X
m3∼M3
ϕ
m1ψ
m2e
X m
α11m
α22m
−α3 2M
1α1M
2α2M
3−α2,
which are general forms of (1.4). The following results will be used in the
proof of Theorem 1.
Corollary 1. Let α
j∈ R with α
1α
2(α
2+1)(α
1−jα
2−j) 6= 0 (j = 1, 2), X > 0, M
j≥ 1, |ϕ
m1| ≤ 1, |ψ
m2| ≤ 1 and let Y := 2 + XM
1M
2M
3. If (κ, λ) is an exponent pair , then for any ε > 0,
S
II{(X
4+6κM
19+11κ+λM
28+10κM
34+6κ)
1/(12+16κ)+ X
1/3M
3/4+λ/(12+12κ)1
M
22/3M
31/3+ (X
3M
18M
26M
34)
1/10+ (X
5M
120M
214M
38)
1/22+ X
1/6M
11/12+λ/(12+12κ)1
M
22/3M
31/3+ (XM
1M
22)
1/2+ (X
2M
123M
214M
38)
1/22+ M
1M
2+ X
−1/2M
2M
3+ X
−1M
1M
2M
3}Y
ε. In particular , if X ≥ M
3≥ M
1, then
S
II{(X
186M
1407M
2350M
3186)
1/536(2.8)
+ (X
164M
1385M
2328M
3164)
1/492+ (X
3M
18M
26M
34)
1/10+ (X
5M
120M
214M
38)
1/22+ (XM
1M
22)
1/2}Y
ε,
S
II{(X
13M
115M
222M
34)
1/26+ (X
2M
12M
23M
3)
1/4(2.9)
+ (X
9M
111M
218)
1/18+ (XM
14M
23M
3)
1/4}Y
ε.
P r o o f. If M
30:= X/M
3≤ ε
0, the Kuz’min–Landau inequality implies S
IIX
−1M
1M
2M
3.
Next we suppose M
30≥ ε
0. As before, using Lemma 2.2 of [12] to the sum over m
3and estimating the corresponding error term by Lemma 2.3 there with n = m
1, we obtain
S
IIX
−1/2M
3S+(X
1/2M
2+M
1M
2+X
−1/2M
2M
3+X
−1M
1M
2M
3) log Y, where
S := X
m1∼M1
X
m2∼M2
X
m03∼M30
e
ϕ
m1ψ e
m2ξ
m03
e
e
α
2X m
β11m
β22m
0β32M
1β1M
2β2M
30β2= X
m1∼M1
X
m02∼M20
e
ϕ
m1ξ e
m02e
e
α
2X m
β11m
0β22M
1β1M
20β2,
and β
j:= α
j/(1+α
2) (j = 1, 2), e α
2:= |1+α
2|·|α
2|
−β2, | e ϕ
m1| ≤ 1, | e ψ
m2| ≤ 1,
|ξ
m03
| ≤ 1, M
20:= M
2M
30, e ξ
m02
:= P P
m2m03=m02
ψ e
m2ξ
m03
. By Theorem 2 with (M, N ) = (M
20, M
1) we estimate S to get the first assertion.
In particular taking (κ, λ) = BA
2 16,
46=
1130,
1630yields
S
II{(X
186M
1407M
2350M
3186)
1/536+ (X
164M
1385M
2328M
3164)
1/492+ (X
3M
18M
26M
34)
1/10+ (X
5M
120M
214M
38)
1/22+ (X
82M
1467M
2328M
3164)
1/492+ (XM
1M
22)
1/2+ (X
2M
123M
214M
38)
1/22+ M
1M
2+ X
−1/2M
2M
3+ X
−1M
1M
2M
3}Y
ε=: (H
1+ H
2+ . . . + H
10)Y
ε.
Since X ≥ M
3≥ M
1, we have H
5≤ H
2, H
7≤ H
4, H
j≤ H
6(8 ≤ j ≤ 10) and thus H
5, H
j(7 ≤ j ≤ 10) are superfluous. This proves (2.8).
The last inequality can be proved similarly by using Theorem 7 of [12]
with (M
1, M
2, M
3) = (M
1, 1, M
20). This completes the proof.
Corollary 2. Let α
j∈ R with α
1α
2(α
1−1)(α
1−2)(α
2+1)(α
1−α
2−1) 6= 0, X > 0, M
j≥ 1, |ψ
m2| ≤ 1 and let Y := 2 + XM
1M
2M
3. If M
3≥ M
1, then for any ε > 0 we have
S
I{(X
7M
18M
210M
36)
1/16+ (X
5M
112M
210M
36)
1/16(2.10)
+ (XM
13M
22M
32)
1/4+ (X
3M
19M
27M
34)
1/11+ (X
2M
19M
26M
34)
1/10+ (X
17M
124M
230M
310)
1/40+ (X
5M
110M
210M
34)
1/14+ (XM
1M
22)
1/2+ X
−1M
1M
2M
3}Y
ε, S
I{(X
15M
111M
222M
34)
1/26+ (X
2M
12M
23M
3)
1/4+ (X
3M
23M
3)
1/4(2.11)
+ (X
11M
17M
218)
1/18+ M
2M
3+ X
−1M
1M
2M
3}(log 2Y )
4. P r o o f. As before we may suppose M
30:= X/M
3≥ ε
0and prove
S
IX
−1/2M
3T (2.12)
+ (X
1/2M
2+ M
1M
2+ X
−1/2M
2M
3+ X
−1M
1M
2M
3) log Y, where
T := X
m1∼M1
X
m2∼M2
X
m03∈I3
g(m
1) e ψ
m2ξ
m03
e
e
α
2X m
β11m
β22m
0β32M
1β1M
2β2M
30β2,
I
3:= [c
3(m
1/M
1)
α1(m
2/M
2)
α2M
30, c
4(m
1/M
1)
α1(m
2/M
2)
α2M
30], and β
j, e α
2, e ψ
m2, ξ
m03
are defined as before, c
j= c
j(α
2) are constants, g(m
1) is a monomial with |g(m
1)| ≤ 1. We define e ξ
m02
and M
20in the same way as in the proof of Corollary 1. Exchanging the order of summation, we have
T = X
m02∼M20
ξ e
m02
X
m1∈I1(m02)
g(m
1)e
e
α
2X m
β11m
0β22M
1β1M
20β2,
where I
1(m
02) is a subinterval of [M
1, 2M
1]. Removing g(m
1) by partial
summation and estimating the double sum obtained by Theorem 3 with
(M, N ) = (M
20, M
1), we find
T {(X
5M
18M
2010)
1/16+ (X
3M
112M
2010)
1/16+ (XM
13M
202)
1/4+ (X
3M
118M
2014)
1/22+ (XM
19M
206)
1/10+ (X
7M
122M
2026)
1/36+ (XM
15M
205)
1/7+ M
11/2M
20}Y
ε.
Inserting into (2.12) and noticing that the last four terms on the right- hand side of (2.12) can be absorbed by (XM
1M
22)
1/2, we obtain (2.10). The inequality (2.11) is (2.7) of [13] with (M
1, M
2, M
3) = (M
2, M
1, M
3) and (α
1, α
2, α
3) = (α
2, α
1, −α
2). This concludes the proof.
3. Proof of Theorem 1. We shall prove only (3.1) ∆(1, 1, 2; x)
εx
47/131+ε,
since this implies ∆(x)
εx
47/131+εby a simple convolution argument. For this we recall some standard notations. Let u := (u
1, u
2, u
3) be a permuta- tion of (1, 1, 2) and let N := (N
1, N
2) ∈ N
2. We write ψ(t) := {t} − 1/2 ({t}
is the fractional part of t) and define S(u, N; x) := X
1
ψ((x/(n
u11n
u22))
1/u3), where the summation condition of P
1
is n
u11n
u22+u3≤ x, n
1(≤)n
2, n
1∼ N
1, n
2∼ N
2. The notation n
1(≤)n
2means that n
1= n
2for u
1< u
2, and n
1< n
2otherwise. It is well known that for proving (3.1) it suffices to verify
S(u, N; x) x
47/131+εfor u = (1, 1, 2), (2, 1, 1), (1, 2, 1).
Since S(1, 1, 2, N; x) x
5/14+ε(see [10], p. 263), it remains to consider u = (2, 1, 1), (1, 2, 1). We shall prove the desired estimate for u = (2, 1, 1) in two cases according to the size of N
1, which we shall formulate as two lemmas. The case of u = (1, 2, 1) can be treated similarly (more easily).
We recall that we have N
1≤ N
2≤ G := x/(N
12N
2), N
1N
2≤ x
1/2when u = (2, 1, 1). This fact will be used (implicitly) many times in the proofs of Lemmas 3.1 and 3.2.
Lemma 3.1. For u = (2, 1, 1), we have
S(u, N; x)
ε{(x
186N
135)
1/536+ (xN
12)
1/4+ (x
40N
17)
1/116+ x
5/14}x
ε. In particular , if N
1≤ x
118/655, then S(u, N; x)
εx
47/131+ε.
P r o o f. By Lemma 2.5 of [12], we have, for any H ≥ 1, (3.2) S(u, N; x) H
−1N
1N
2+ (log x) max
1≤H0≤H
H
0−1|S(H
0, N)|, where
S(H
0, N) := X
h∼H0
a
hX
n1∼N1
X
n2∼N2
e(hx/(n
21n
2)), |a
h| ≤ 1.
The inequalities (2.8) and (2.9) with (X, M
1, M
2, M
3) = (GH
0, N
1, H
0, N
2) imply
S(H
0, N) {(G
186N
1407N
2186)
1/536+ (GN
1H
0)
1/2+ χ
1+ χ
2}H
0x
ε, S(H
0, N) {(G
13N
115N
24H
09)
1/26+ (G
2N
12N
2H
0)
1/4+ (G
9N
111H
09)
1/18+ (xN
12)
1/4}H
0x
ε=: {D
1+ D
2+ D
3+ (xN
12)
1/4}H
0x
ε,
with χ
1:= (G
3N
18N
24H
0−1)
1/10and χ
2:= (G
5N
120N
28H
0−3)
1/22, where we have used the fact that (G
164N
1385N
2164)
1/492≤ (G
186N
1407N
2186)
1/536(in view of N
1≤ x
1/4). From these, we deduce that for any H
0≥ 1,
S(H
0, N) n
(G
186N
1407N
2186)
1/536+ (GN
1H
0)
1/2+ (xN
12)
1/4+ X
1≤j≤2
X
1≤k≤3